Mathematics - 2017-01-23 (page 4 of 4)

Well I have the proof with complex matrices.
I was wondering if it held that since two real matrices that are similar in $M_n(\Bbb C)$ are similar in $M_n(\Bbb R)$ and since $M_n(\Bbb R)$ is closed in $\Bbb M_n(\Bbb C)$, then it follows that $Sim(M)$ is closed iff $M$ is diagonalizable in $M_n(\Bbb R)$, but I don't think that's enough.
(By the way, is diagonalizable the right term ? It seems really weird)

I guess it's true but to prove it one has to do the same proof as in $M_n(\Bbb C)$ all over again

Semiclassical

10:12 PM

Which I guess goes with what you said above, since you can write $\mathbf{G}=(d+\mathcal{A})\wedge \mathcal{A}$?

Astyx

@Ted I don't think I have, at least I don't know the name

Ted Shifrin

So how do you know non-diagonalizable implies not closed?

Semiclassical

Oh, I do have an actual question for you.

Secret

Aren't all square matrices similar to their Jordan decomposition?

Ted Shifrin

Over an algebraically closed field?

@Semiclassic, actually the curvature should be $(d+\mathcal A)^2$, suitably interpreted.

Semiclassical

10:15 PM

Hmm.

Secret

I mean, any square matrix can always be written as a block triangular matrix sandwiched between unitary matrices?

Semiclassical

Suppose I have (as I said earlier, though with slightly different notation and in less generality) the affine curve $f(x,y)=(y-x)(x-x_1)(x-x_2)(x-x_3)=0$.

Astyx

I have a proof where you take $D = Diag(1,2,\dots, n)$ and take $A$ triangular (which you can without leaving the similarity class in $M_n(\Bbb R)$) and then compute $D^kAD^{-k}$. This converges to $Diag(a_{1,1}, \dots, a_{n,n})$ which is by hypothesis in $Sim(A)$. The result follows

I don't think we're going to study Jordan canonical form this year

Semiclassical

As a projective variety, there'll be 3 Riemann spheres with a common (triple) point and another Riemann sphere which shares double points with the others.

Ted Shifrin

Where are you using non-diagonalizable, then, @Astyx?

Right, @Semiclassic.

Semiclassical

10:18 PM

First question: How can a triple point resolve?

Astyx

I proved the contrapositive @Ted

Semiclassical

With a double point I expect to replace the node with a handle.

Ted Shifrin

To think about the converse, I think you need Jordan form.

Semiclassical

But I'm not sure what can happen with a triple point.

Ted Shifrin

Right, @Semiclassic. You're unpinching a pinched torus.

Semiclassical

10:19 PM

Right.

Ted Shifrin

Triple point comes from higher genus.

But if you wiggle it you should get 3 nodes nearby.

So it's like starting with 6 nodes.

Astyx

The converse can be solved using minimal and caracteristic polynomials

Semiclassical

Hmm.

Ted Shifrin

Understanding minimal polynomials is a good part of Jordan form, @Astyx.

Semiclassical

I'm not really following what is meant by 'wiggling.'

Though I presume it amounts to a certain perturbation.

Ted Shifrin

10:21 PM

Generic small deformation.

Astyx

Then when $A$ is diagonalizable, it follows that $Sim(A) = \chi^{-1}(\{\chi_A\})\cap \mu_A^{-1}(\{0\})$

Ted Shifrin

I have no idea what that means, @Astyx.

Semiclassical

Oh, I think I see what you mean (maybe).

I start with a triple point, but if I wiggle then I can replace that triple point with another Riemann sphere which touches the others in nodes.

And then if I perturb that, I get handles.

Astyx

Oh sorry, $\chi : A \mapsto \chi_A$ is the function that maps a matrix to its caractristic polynomial, and $\mu_A$ is the minimal polynomial of $A$. And $Sim(A)$ the similarity class of $A$

I might be using Jordan form without knowing it

Semiclassical

The difficulty in my case is that my perturbation is not generic: $f(x,y)\to f_t(x,y)=(y-x)(x-x_1)(x-x_2)(x-x_3)-t(x-x_1)(x-x_2)=0$.

Ted Shifrin

10:24 PM

It still doesn't make sense. You need functions on the space of matrices. So what you're writing down isn't right.

Semiclassical

(I really have 3 such perturbations, but their coefficients are independent so I wanted to start with that one.)

Astyx

Do you mean me @Ted ?

Ted Shifrin

Yes.

Astyx

My functions are on the space of matrices, are they not ?

Semiclassical

What I want to say is that I can rewrite the perturbed curve as $[(y-x)(x-x_3)-t](x-x_1)(x-x_2)=0$.

Ted Shifrin

10:27 PM

So just as a double point drops genus by 1, a triple point drops it by 3.

Semiclassical

Problem is, I don't know how that affects the triple point.

Ted Shifrin

@Astyx: I think you mean $\mu^{-1}(\{\mu(A)\})$ or something.

Well, write than down carefully in homogeneous coordinates, @Semisimple.

Semiclassical

Heh.

Fair enough.

Ted Shifrin

The $t$ needs a $z^2$.

Semiclassical

Well, I'm only writing down the affine curve.

If I want the projective one, the x_k's also need z's.

Ted Shifrin

10:29 PM

Yes.

Semiclassical

What I really want, at the end of the day, is three such perturbations.

Ted Shifrin

Hold on and figure out one.

Astyx

$\mu$ is not continuous, so that wouldn't work. I mean $$\mu_A : \begin{cases}M_n(\Bbb K) \mapsto M_n(\Bbb K)\\B \mapsto \mu_A(B)\end{cases}$$

Semiclassical

Yeah.

Ted Shifrin

OK, I cannot multitask any more.

Astyx

10:31 PM

I'll leave then :)

I need to sleep anyways

Bye !

Semiclassical

Main reason I mention that is because I can write the general case as $$y=x+\frac{t_1}{x-x_1}+\frac{t_2}{x-x_2}+\frac{t_3}{x-x_3}$$

Astyx

And thanks

Ted Shifrin

Bye.

Semiclassical

(or as many as I want, really).

Ted Shifrin

Semiclassic. Quit it. Focus.

That perturbation doesn't move anything at infinity.

Balarka Sen

10:33 PM

@Semiclassical A link of the triple point looks like a Hopf linkage. Three circles, each linked to each other by linking number 1.

Semiclassical

Hmm, then maybe I'm making things more complicated than I should.

Balarka Sen

You cut out the singular interior and glue back it's Siefert surface.

Ted Shifrin

Does that explain why normalization adds 3 to the genus?

its

Balarka, of course we're trying to stay in the algebraic category here ...

Balarka Sen

What I say can be done algebraically, I am quite sure.

Ted Shifrin

Well, I'm resolving the triple point into three nodes ... then resolving them.

Balarka Sen

10:35 PM

Because what you do for a node is exactly what I say; cut out the singular interior of the Hopf link and glue back it's Siefert surface. That's going from $z_1z_2 = 0$ to $z_1z_2 = \epsilon$.

Semiclassical

Hmm, it looks like $(y-x)(x-x_1 z)-tz^2=0$ has no singularities (for $t\neq 0$.)

Balarka Sen

@TedShifrin That's also a possible thing to do. Do we end up with the same genus?

Yours add 3...

Ted Shifrin

Right, and that's what a triple point should do. Do you see it your way?

Semiclassical

you = ?

Ted Shifrin

You should learn about Milnor fibers. There's another cool Milnor book for you to read, Balarka.

Balarka Sen

10:37 PM

What's the genus of the Siefert surface of three circles with each pair linked by linking number 1? I wonder if it's 3.

Ted Shifrin

That's too much topology for me :P

Semiclassical

I'm okay with not talking about milnor fibers atm :)

Ted Shifrin

I'm talking to Balarka, not you.

Balarka Sen

@TedShifrin Ah, yeah, I have heard of it.

Semiclassical

Hence why I asked "you = ?" :P

Ted Shifrin

10:39 PM

I'm just saying that in your original, setting $z=0$ kills the $t$ term and so it does us no good for changing the triple point. Or am I missing something, @Semiclassic?

Semiclassical

No, you're probably right.

Kasmir Khaan

hi chat , how to find the intersection of 2 surfaces in space ? am doing stokes theorem

Semiclassical

So I shouldn't expect this perturbation to affect the singularities at infinity whatsoever (?).

Ted Shifrin

Too vague a question, Kasmir.

Kasmir Khaan

ill write on latex and post an example

Alessandro Codenotti

10:41 PM

is there a name for an inner product which is positive semidefinite instead of positive definite?

Ted Shifrin

Well, the point $[0,1,0]$ is still on those curves for all $t$, @Semiclassic.

Semiclassical

Hm.

Ted Shifrin

usually a degenerate inner product, @Alessandro. I dunno.

Semiclassical

One other thing I'm interested in, and this is where I get confused, is that if I consider the one-form $y\,dx$

Kasmir Khaan

$x^2+y^2+z^2 =1$ and $x+y+z=0$ , $z=-x-y$ --> $x^2+y^2+xy =1/2$

Ted Shifrin

10:42 PM

I do examples like that in my lectures, @Kasmir.

Semiclassical

then when I've got $y=x+\frac{t}{x-x_1}$ then it'll have a pole at $x=x_1$ whenever $t\neq 0$.

Ted Shifrin

Don't do so much algebra. You don't want to project to the $xy$-plane. You want to work in the plane $x+y+z=0$. @Kasmir

Kasmir Khaan

hmm but i dont get the concept of that

Semiclassical

(I, by contrast, probably shouldn't be projecting onto the xy plane so much.)

Kasmir Khaan

how does the condition of unit sphere play roll here?

Ted Shifrin

10:43 PM

Depends on the problem, @Kasmir, but I gave my students lots like that where you should just figure out $\text{curl} F\cdot n$ using the plane.

Kasmir Khaan

if all i care about is the plane $x+y+z=0$

okay ill watch that lecture now

Ted Shifrin

It intersects the plane in a great circle. But that circle bounds a disk in the plane.

Semiclassical

wha-oh, gotta catch the train

Ted Shifrin

Bye, @Semiclassic. I'm not following you anyhow.

Kasmir Khaan

you use differentials to deal with surface integrals and you asked me not to do it that way

so i did not watch further =p

but I assume that strokes theorem cant be taught in one way right?

Ted Shifrin

10:45 PM

I talked about interpreting it as a flux directly. What's your $F$?

Kasmir Khaan

F = ( 3y-2z , z-3x , 2x-y )

Ted Shifrin

Right. Compute curl. It is a constant vector field. How do you calculate the flux across a region in the tilted plane? $\text{curl}\,F\cdot n$ will be a CONSTANT.

Balarka Sen

@TedShifrin Ya, it's 3.

Kasmir Khaan

Il ldo the curl now

but that is not where I have difficulty

its the set up again

Ted Shifrin

Had to be, @Balarka. I have no idea why, but it had to be :)

I'm saying that you need to think, @Kasmir, not grind formulas.

Kasmir Khaan

10:48 PM

I mean we have two surfaces and they intersect to form a closed curve

Ted Shifrin

What's the integral of a constant over a circular region of radius $1$?

Kasmir Khaan

the area of it

pi

Ted Shifrin

times the constant. Right. Done.

No horrible formulas. Just understanding what you're doing with the theorem.

Semiclassical

(Caught the train in time, good thing I hurried)

Ted Shifrin

IT's an excellent sort of homework and test question.

LOL @Semiclassic. You couldn't run all the way home?

Balarka Sen

10:49 PM

What does degree-genus say about it? 3 is the minimal possible genus, yeah?

I forgot the formula

Semiclassical

Not when I'm living with parents in the suburbs.

I take a 20 minute light rail trip to St Paul, then a half hour bus ride to the northern suburbs

Can't run that :p

Kasmir Khaan

@TedShifrin how do you know that the plane intersects the sphere in a circular shape algebraiclly ?

because we deal also with many other surfaces that are hard to see

Ted Shifrin

You don't see it algebraically unless you use linear algebra to change coordinates. You use geometry instead.

Kasmir Khaan

hmm okay =p so if I understood correctly now

r (x,y) = (x , y , -x-y )

Semiclassical

Anyways. I suspect what I don't grok right now is how contour integrals over a projective curve in CP^2 and over the affine curve in C^2 are related.

I ultimately am interested in integrals on the curve rather than the curve itself.

Ted Shifrin

10:55 PM

Don't do that, @Kasmir. You do NOT want to do it algebraically. You have to do it by thinking.

Semiclassical

But I'm feeling pretty handwavy now

Kasmir Khaan

hmm am absoluley lost , the plane intersect the sphere in a circle right?

i need that information about circle becuse am gonna do surface integral

of that region

Ted Shifrin

@Semiclassic: Your affine curve is the whole curve except for a finite number of points. The question is whether your 1-form has singularities there?

Read all the things I've said to you for the last 20 minutes, @Kasmir.

Kasmir Khaan

Ok I will , thanks :)

Semiclassical

The other thing I'm getting mixed up on in practice are singularities of the curve versus those of the one-form

Ted Shifrin

11:00 PM

The singularities of the curve shouldn't actually affect integrating the 1-form. If the path happens to go through a singular point, you pull back to a desingularization of the curve and you have no problem. I guess there are issues with what the 1-form means at such points.

Semiclassical

Hmm.

Ted Shifrin

Aren't you asleep, @Alessandro? :D

Kasmir Khaan

@TedShifrin so far am here double integral of surface , ( -2 , 4, -6 ) dot ( 1,1,1 ) normal of the plane

Balarka Sen

neither am I

Ted Shifrin

UNIT normal, @Kasmir, UNIT normal.

Semiclassical

11:04 PM

I'm insisting on doing this in a curve-centric way because ultimately I'll only have $y(x)$ by virtue of $f(x,y)=0$.

Kasmir Khaan

@TedShifrin oh yes >< sqrt3 , okay but ehm now my question is what is that surface

Alessandro Codenotti

I'm about to go to sleep since the numerical analysis exam begins in 9 hours or so

Good night everyone

Kasmir Khaan

i know its a circle but what radius

Ted Shifrin

It's the inside of the circle of radius 1 in the plane.

Balarka Sen

night

Ted Shifrin

11:06 PM

It's radius 1 because the plane goes through the origin. If the plane didn't go through the origin, you'd need to use the Pythagorean Theorem to find its radius.

Kasmir Khaan

how do you know that?

Ted Shifrin

Draw a picture and think.

Balarka Sen

hmm, I should read some of Ginsberg at some point.

Semiclassical

Ok, out for now to avoid getting a headache from typing on mobile

Kasmir Khaan

@TedShifrin i see what you mean now

@TedShifrin because it goes through the origin it cuts the sphere into 2 equal pieces

but diagonally

Ted Shifrin

11:08 PM

If I gave you the plane $x+y+z=1/2$, what would be the radius of the circle?

Kasmir Khaan

let me think

we shifted the plane by a 1/2

hmm I dont know how to do this geometrilly

but i assume i need to solve it for z and then plug in on the unit sphere

Ted Shifrin

Draw a 2-dimensional cross-section of the picture. Circle, line ... How far is the line from the center of the circle?

Kasmir Khaan

r = 1/4

Ted Shifrin

Hold on. You jumped to the end.

Kasmir Khaan

@TedShifrin the way i did it was to draw it on yz plane and found the y intersect

Ted Shifrin

11:23 PM

I don't think that's right. The closest point to the origin on the plane won't be on the yz plane.

Kasmir Khaan

hmm why do we need to find the closest point ?

i found the diameter

and took half of it

Ted Shifrin

Because the center of the circle will be the point on the plane closest to the center of the sphere (the origin).

Kasmir Khaan

y = 1/4 + sqrt (7/4) , 1/4 - (sqrt7) /4

Ted Shifrin

No. Remember the normal to the plane is $(1,1,1)/\sqrt{3}$.

Kasmir Khaan

Can you show me how you thinking ?

Ted Shifrin

11:30 PM

I want to find how far along $n=(1,1,1)/\sqrt3$ I have to travel to meet the plane. That point will be the center of the circle I'm looking for.

So I get a little right triangle — the distance I traveled and the radius of the circle are the legs of the right triangle, and a radius of the sphere (length 1) is the hypotenuse.

Kasmir Khaan

You use pytagoras ?

Ted Shifrin

Eventually, yes.

But first you want to know for what value of $t$ you have $t(1,1,1)/\sqrt 3$ lying on the plane $x+y+z=1/2$.

Kasmir Khaan

hmm thats a smarter way of doing it

Ted Shifrin

Part of what you need to learn is to visualize things and try to use concepts, not always horrible calculations.

Kasmir Khaan

well we did not do much of space geometry

mt3

11:33 PM

hello can somebody please tell me what the dimension of an open cover is?

Kasmir Khaan

so I have absolut no backround on it

Ted Shifrin

No time like the present, @Kasmir :)

Kasmir Khaan

@TedShifrin I see how that would be usefull in the theorems am using but do you have videos from basic ?

we started with those theorems in space without any knowledge of space geometry

thats why I got lost

Ted Shifrin

@mt3: Are you talking about covering dimension? You can read on Wikipedia, I'm sure.

mt3

oh i searched for it, but the term "Lebesgue" in the results put me off

Ted Shifrin

11:36 PM

@Kasmir: There were lectures at the beginning of the course on projections and thinking about level curves and level surfaces. I don't think I ever did explicitly what we were just doing.

mt3

but apparently that is what i was looking for

thanks

Ted Shifrin

@mt3: He's the one who came up with it.

Kasmir Khaan

@TedShifrin okay thanks alot Ted! you are the only person who can get this ideas into my head :D

Ted Shifrin

But you should try to draw a picture looking at the plane on the edge ... so the normal is perpendicular and then you see the triangle we were just talking about. @Kasmir.

You should try discussing stuff with more of your friends taking the course, @Kasmir. Working together and teaching one another is a good thing.

Kasmir Khaan

@TedShifrin We do that :)

Ted Shifrin

11:38 PM

Need to draw more pictures!! And make sure they make sense.

Kasmir Khaan

on some problems we have to find r (s,t) = (s,t,f(s,t))

and project the 3 d surface into st plane

and sometimes the geometry is too hard so that is why am trying to do all algebraiclly

Ted Shifrin

You need to do lots of problems and figure out when to do what. Just like you did when you first learned to take integrals ...